How do you find the integral #ln(x) x^(3/2) dx#?

1 Answer
Jun 18, 2015

To do integration by parts:

#uv = intvdu#

I would pick a #u# that gives me an easy time differentiating, and a #dv# that I would easily be able to integrate multiple times if necessary. If the same integral comes back, or variables aren't going away, either it's cyclic or you should switch your #u# and #dv#.

Let:
#u = lnx#
#du = 1/xdx#
#dv = x^(3/2)dx#
#v = 2/5x^(5/2)#

#=> 2/5x^(5/2)lnx - int2/5(x^(5/2)/x)dx#

Nice! That's not too bad.

#= 2/5x^(5/2)lnx - int2/5x^(3/2)dx#

#= 2/5x^(5/2)lnx - 4/25x^(5/2) + C#

#= 10/25x^(5/2)lnx - 4/25x^(5/2) + C#

#= x^(5/2)[10/25lnx - 4/25] + C#

#= 2/25x^(5/2)[5lnx - 2] + C#